JEE Main 2021 · 26 February · morning · Single correct · +4 / −1

The maximum value of the term independent of 't' in the expansion
of ( tx^(1/5) + (1 - x)^(1/10)/t )¹⁰ where x∈(0, 1) is :

  1. (a) 10!/(√3 (5!)²)
  2. (b) 2.10!/(3√3 (5!)²)
  3. (c) 10!/3(5!)²
  4. (d) 2.10!/3(5!)²
Answer and solution

Answer (b) 2.10!/(3√3 (5!)²)

Solution

Maths sir · 4 steps

Make the powers of t cancel to fix r = 5, then maximise what is left, x√(1-x), using calculus.

  1. General term. The (r+1)th term is C(10, r)(t x^(1/5))¹⁰⁻ʳ((1-x)^(1/10)/t)ʳ. The power of t is (10-r)-r = 10-2r, which is zero only when r = 5.

    T₆ = C(10, 5) x^(5/5) (1-x)^(5/10) = 10!/(5!)² x√(1-x)

  2. Maximise the x part. Let f(x) = x√(1-x). It is easier to maximise g(x) = f(x)² = x² - x³. Then g'(x) = 2x - 3x² = 0 gives x = 2/3, and g''(x) = 2 - 6x = -2 < 0 there, so it is a maximum.
  3. Value at the maximum. f(2/3) = 2/3√(1/3) = 2/3√3.
  4. So. The maximum term independent of t is 10!/(5!)² × 2/3√3 = (2 × 10!)/(3√3 (5!)²), which is option (b).

The idea to keep. For a term independent of a variable, set its total power to zero to find r. If the leftover depends on another variable, maximise it with calculus. Squaring first removes the root.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2021.